Problem 91 - A Complete Resource Book in Chemistry for JEE Main

Question

Heating methyl phenyl ether with HI gives (a) methanol + iodobenzene (b) methyl alcohol + benzyl alcohol (c) methyl iodide \(+\) phenol (d) methyl iodide + iodobenzene

Step-by-Step Solution

Answer

(c) methyl iodide + phenol

Step 1: Understanding the reaction

The reaction we are analyzing involves heating methyl phenyl ether with hydroiodic acid (HI). Methyl phenyl ether is also known as anisole, which has the structure C6H5-O-CH3. In the presence of HI, this ether undergoes nucleophilic substitution.

Step 2: Identifying the leaving group

In the nucleophilic substitution reaction, HI will cause the bond between the oxygen and one of the carbon atoms to break. The iodide ion from HI is a good nucleophile, and it attacks the more hindered carbon atom of the ether, which in this case is the methyl group (CH3).

Step 3: Breaking the ether bond

The ether bond will cleave, favoring the formation of a more stable iodide salt. Therefore, the oxygen-CH3 bond will break, resulting in methyl iodide (CH3I) and the leaving group will most likely remain on the phenyl portion, leading to the formation of phenol (C6H5OH), not iodobenzene.

Step 4: Writing the final products

After the bond cleavage, we obtain methyl iodide (CH3I) and phenol (C6H5OH) as the main products of the reaction. It matches choice (c) from the options given in the problem.

Key Concepts

Ether Cleavage

Ether cleavage is a common reaction in organic chemistry where an ether is split into two parts. This typically involves breaking the carbon-oxygen bond. When a reagent like hydroiodic acid (HI) is used, it provides an iodide ion, which helps in cleaving the ether. In our case, methyl phenyl ether (anisole) undergoes this reaction.

• The ether bond being targeted is between oxygen and one of the carbon atoms.
• This reaction is a type of nucleophilic substitution because the nucleophile (iodide ion) attacks a carbon atom in the ether.
• Cleavage results in the formation of different products depending on the specific structure of the ether involved.

Understanding the specifics of the ether structure, in this case, helps predict the products of the reaction effectively.

Iodide Ion Nucleophile

In nucleophilic substitution reactions, the iodide ion often plays a crucial role as a nucleophile. A nucleophile is a chemical species that donates an electron pair to an electrophile to form a chemical bond in a reaction. Due to its negative charge and size:

• The iodide ion is very effective at attacking carbon atoms.
• It tends to favor reactions where it can disrupt existing bonds, leading to new product formation.
• In the reaction with anisole, the iodide ion attacks the more hindrances-free methyl group.

Its role as a good nucleophile means it excels in promoting ether bond cleavage to form products like methyl iodide.

Methyl Iodide

Methyl iodide, or CH3I, is one of the products formed during the ether cleavage of methyl phenyl ether with HI. When discussing methyl iodide:

• It forms when the iodide ion successfully breaks the bond between the ether oxygen and the methyl group.
• Given its simple structure, methyl iodide is a typical primary alkyl halide.
• This compound is often used in organic synthesis as a methylating agent.

The formation of methyl iodide highlights the effective role of iodide ions in nucleophilic substitutions.

Phenol Formation

Phenol is an important compound formed during the reaction of anisole with HI. When the ether bond in methyl phenyl ether is cleaved, phenol forms:

• As the other product when the methyl group leaves as methyl iodide.
• The hydroxyl group remains attached to the phenyl ring to form phenol, C6H5OH.
• Phenol has relevance in various applications, from industrial processes to pharmacology.

This transformation illustrates how ether cleavage can be utilized to form useful aromatic alcohols like phenol in organic reactions.